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x(x+16)=35
We move all terms to the left:
x(x+16)-(35)=0
We multiply parentheses
x^2+16x-35=0
a = 1; b = 16; c = -35;
Δ = b2-4ac
Δ = 162-4·1·(-35)
Δ = 396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{11}}{2*1}=\frac{-16-6\sqrt{11}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{11}}{2*1}=\frac{-16+6\sqrt{11}}{2} $
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